标签：des style color os io strong for ar
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 9963 | Accepted: 4179 | Special Judge | ||
Description
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:
Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.
Input
On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).
Output
The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.
Sample Input
3 5 4
Sample Output
6 FILL(2) POUR(2,1) DROP(1) POUR(2,1) FILL(2) POUR(2,1)
和之前做过的一道二维bfs一样，只不过这个需要回溯路径，很简单在结构体中加一个变量来记录上一个状态在队列中的下标（手敲的队列比较好，这个时候在用STL队列好像不大方便）最后找到满足条件的状态逆向打印路径（因为记录的都是上一个状态）
#include <cstdio> #include <iostream> #include <cstring> #include <cstdlib> #include <queue> using namespace std; int m,n,c,s,e,p; typedef struct node { int v1,v2,cur,pre,op; }; bool vis[999][999]; int ans[10010]; node que[10010]; void bfs() { p=0;s=0;e=0;int pos; node t={0,0,0,0,0}; que[e++]=t; vis[0][0]=1; while(s<e) { node f=que[s];pos=s;s++; if(f.v1==c||f.v2==c) { printf("%d\n",f.op); int tem=pos; for(int i=0;i<f.op;i++) { ans[p++]=que[tem].cur; tem=que[tem].pre; } for(int i=p-1;i>=0;i--) { switch(ans[i]) { case 1:printf("FILL(1)\n");break; case 2:printf("FILL(2)\n");break; case 3:printf("DROP(1)\n");break; case 4:printf("DROP(2)\n");break; case 5:printf("POUR(2,1)\n");break; case 6:printf("POUR(1,2)\n");break; } } return ; } if(f.v1!=m) { t.v1=m; t.op=f.op+1; t.v2=f.v2; if(!vis[t.v1][t.v2]) { vis[t.v1][t.v2]=1; t.cur=1; t.pre=pos; que[e++]=t; } } if(f.v2!=n) { t.v2=n; t.op=f.op+1; t.v1=f.v1; if(!vis[t.v1][t.v2]) { vis[t.v1][t.v2]=1; t.cur=2; t.pre=pos; que[e++]=t; } } if(f.v1!=0) { t.v1=0; t.v2=f.v2; t.op=f.op+1; if(!vis[t.v1][t.v2]) { vis[t.v1][t.v2]=1; t.cur=3; t.pre=pos; que[e++]=t; } } if(f.v2!=0) { t.v2=0; t.v1=f.v1; t.op=f.op+1; if(!vis[t.v1][t.v2]) { vis[t.v1][t.v2]=1; t.cur=4; t.pre=pos; que[e++]=t; } } if(f.v2!=0&&f.v1!=m) { t.v2=f.v2-(m-f.v1);if(t.v2<0) t.v2=0; t.v1=f.v1+f.v2; if(t.v1>m) t.v1=m; t.op=f.op+1; if(!vis[t.v1][t.v2]) { vis[t.v1][t.v2]=1; t.cur=5; t.pre=pos; que[e++]=t; } } if(f.v1!=0&&f.v2!=n) { t.v1=f.v1-(n-f.v2);if(t.v1<0) t.v1=0; t.v2=f.v2+f.v1; if(t.v2>n) t.v2=n; t.op=f.op+1; if(!vis[t.v1][t.v2]) { vis[t.v1][t.v2]=1; t.cur=6; t.pre=pos; que[e++]=t; } } } puts("impossible"); } int main() { while(cin>>m>>n>>c) { memset(vis,0,sizeof(vis)); bfs(); } return 0; }
POJ 3414--Pots--BFS+回溯路径,布布扣,bubuko.com
标签：des style color os io strong for ar
原文地址：http://blog.csdn.net/qq_16255321/article/details/38546341
